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Suppose two points are situated at the origin of xy coordinate system. Two points start moving at the same time. Point $T_1$ starts moving towards north with velocity $v_1 = 2.7m/s$ while point $T_2$ starts moving towards east with velocity $v_2 = 1.6 m/s$ and acceleration $a_2 = 0.9 m/s^2$.
What is their relative velocity at time $t = 10s$?

1. solution:
We can describe point $T_1$ with vector $\vec{v_1} = v_1\hat{j}$ and point $T_2$ with vector $\vec{v_2} = (v_2+a_2t)\hat{i}$
If we now subtract $\vec{v_2}$ from $\vec{v_1}$ we get $\vec{v_r} = v_1\hat{j}-(v_2+a_2t)\hat{i}$
It's length is $\sqrt{v_1^2+(v_2+a_2t)^2}$

2. solution:
We can calculate the distance between two points using pythagorean theorem: $d=\sqrt{(v_1t)^2+(v_2t+\frac{a_2t^2}{2})^2}$
$\textbf{It's derivative is relative velocity.}$

If we now graph those two functions, we don't get the same graph:enter image description here

Why is there a diference between those two methods?

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  • $\begingroup$ Actually, In (2) I calculate derivative of distance with repsect to time. Why wouldn't change in distance with respect to time be their relative velocity? $\endgroup$ Mar 8, 2020 at 18:37
  • $\begingroup$ I sure do not understand your point. The first is a velocity. But in the second method I calculate the distance so I can calculate the derivative of distance, which is velocity. There is surely something I am doing wrong, but I don't see why change in distance with repsect to time wouldn't be their relative velocity. $\endgroup$ Mar 8, 2020 at 18:49
  • $\begingroup$ This question should not be closed as homework-like because it actually involves an important conceptual issue. $\endgroup$
    – G. Smith
    Mar 8, 2020 at 19:41

2 Answers 2

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The distance between two points... it’s derivative is relative velocity.

No, it isn’t. The derivative of the separation distance is the component of the relative velocity along the line between the two objects, not the relative speed (i.e., the magnitude of the relative velocity).

$$\frac{d|\mathbf{r}_1-\mathbf{r}_2|}{dt}=\frac{\mathbf{r}_1-\mathbf{r}_2}{|\mathbf{r}_1-\mathbf{r}_2|}\cdot(\mathbf{v}_1-\mathbf{v}_2)\ne|\mathbf{v}_1-\mathbf{v}_2|$$

The component of the relative velocity that is perpendicular to the separation does not contribute to the instantaneous rate of increase of the separation distance. For example, in the case of two objects in circular orbits about their center of mass, they have a relative velocity but there is no change in their separation distance.

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  • $\begingroup$ I apologize about my ignorance. This is an amazing answer. This would work if points are moving along the same line but not otherwise. Thank you for the answer. $\endgroup$ Mar 8, 2020 at 19:57
  • $\begingroup$ No need for apologies. This is a common misunderstanding. $\endgroup$
    – G. Smith
    Mar 8, 2020 at 20:42
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In the first case:

$$\mathbf v_1 - \mathbf v_2 = \frac{ d\mathbf r_1}{dt} - \frac{d\mathbf r_2}{dt} = \frac{d(\mathbf r_1 - \mathbf r_2)}{dt}$$

In the second case:

$$\frac{d|\mathbf r_1 - \mathbf r_2|}{dt}$$

Clearly they are different. The second one measures how the distance between the observers changes with time, or the radial velocity, not considering the tangential velocity.

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